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We can extend the technique of trigonometric substitution to the hyperbolic functions. Use Theorem 2.20 and the identity cosh2x-sinh2x=1to solve each integral in Exercises 87– 90 with an appropriate hyperbolic substitution x=asinhuorx=acoshu.. (These exercises involve hyperbolic functions.)

1x2+4dx

Short Answer

Expert verified

The value of the integral issinh-1x2+C.

Step by step solution

01

Step 1. Given information.

The given integral is1x2+4dx

02

Step 2. Substitution.

To solve the integral,

Letx=sinh(u)dx=cosh(u)du

sinh2(u)+1=cosh2(u).

Now,

1x2+4=14sinh2(u)+4=12sinh2(u)+1=12cosh2(u)=12cosh(u)

03

Step 3. Value of the integral.

Now, we can write,

1x2+4dx=1du=u=sinh-1x2+C

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